Answer:
Explanation:
Height of building
H = 6m
Horizontal speed of first balloon
U1x = 2m/s
Second ballot is thrown straight downward at a speed of
U2y = 2m/s
Time each gallon hits the ground
Balloon 1.
Using equation of free fall
H = Uoy•t + ½gt²
Uox = 0 since the body does not have vertical component of velocity
6 = ½ × 9.8t²
6 = 4.9t²
t² = 6 / 4.9
t² = 1.224
t = √1.224
t = 1.11 seconds
For second balloon
H = Uoy•t + ½gt²
6 = 2t + ½ × 9.8t²
6 = 2t + 4.9t²
4.9t² + 2t —6 = 0
Using formula method to solve the quadratic equation
Check attachment
From the solution we see that,
t = 0.9211 and t = -1.329
We will discard the negative value of time since time can't be negative here
So the second balloon get to the ground after t ≈ 0.92 seconds
Conclusion
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
Is there options for this??
Answer:
[1, 6, -2]
Explanation:
Given the following :
Initial Position of spaceship : [3 2 4] km
Velocity of spaceship : [-1 2 - 3] km/hr
Location of ship after two hours have passed :
Distance moved by spaceship :
Velocity × time
[-1 2 -3] × 2 = [-2 4 -6]
Location of ship after two hours :
Initial position + distance moved
[3 2 4] + [-2 4 -6] = [3 + (-2)], [2 + 4], [4 + (-6)]
= [3-2, 2+4, 4-6] = [1, 6, -2]
Answer
given,
Pressure on the top wing = 265 m/s
speed of underneath wings = 234 m/s
mass of the airplane = 7.2 × 10³ kg
density of air = 1.29 kg/m³
using Bernoulli's equation
Applying newtons second law
2 Δ P x A - mg = 0
A = 3.53 m²
Answer:
625 W
Explanation:
Applying
P = W/t.................... Equation 1
Where p = power, W = Work, t = time
But,
W = Force (F) × distance (d)
W = Fd........................ Equation 2
Substitute equation 2 into equation 1
P = Fd/t.................... Equation 3
From the question,
Given: F = 5000 N, d = 30 m, t = 4 munites = (4×60) seconds = 240 seconds
Substitute these values into equation 3
P = (5000×30)/240
P = 625 Watt
